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0

Yes , the normal to the surface is the direction of reaction force. And the direction doesnt depend on the material of the object . But note that if friction is considered , direction of net reaction force changes


2

Because it is a perfectly elastic collision the kinetic energy and the momentum are conserved. So you have two equations for two unknowns which are the final velocity of the football player and his mass: $$ m_f v_f^0+m_r v_r^0=m_f v_f^1+m_r v_r^1 $$ $$ \frac{m_f (v_f^0)^2}{2}+\frac{m_r (v_r^0)^2}{2}=\frac{m_f (v_f^1)^2}{2}+\frac{m_r (v_r^1)^2}{2} $$ and ...


0

In a perfectly elastic collision, the final momentum of the system should be equal to the initial momentum of the system. It seems to be set up correctly, so I would say that you would need additional information for this.


0

Use conservation of momentum, which tells you that the total momentum (the sum of the momenta of the two particles) before and after collision must be the same. Also note that the momentum is a function of the vector velocity, which means that you can make two independent analyses, one on the $x$-axis, and one on the $y$-axis. Both should respect ...


0

Considering both the discs as the system , we can conserve angular momentum about their collinear axis of rotation . The torque due to friction will decrease the angular velocity of the disc having more angular momentum (before the collision ) while the torque will increase angular velocity of the one which had lesser initial angular momentum . I am assuming ...


-1

logically in the frame of centre of mass the accn of body is zero so momentum is conserved and as mass has not changed initial velocity is equal to final velocity in frame of centre of mass


0

Start with the two pertinent conservation laws for elastic collisions: kinetic energy and momentum. Remember that momentum is a vector. In the center of mass frame, the total momentum is zero. That will get you started. Do the work for two particles first. As an aside you should try to show the total momentum is zero in the CoM frame by example by taking ...


0

Can someone explain how two objects with different masses can have the same initial velocity? Since the kinetic energy $E_{kin}=m\cdot v^2/2$ you need more energy = more gunpowder to get the same velocity for a heavier bullet. What would be different about the final state of the apparatus, i.e. the angle of the pendulum? The pendulum would swing ...


0

Yes, it would of course. And so the resultant weight (your's + the weight of what you are carrying) starts acting through this new and horizontally shifted (maybe vertically shifted too) Center of Mass. Notice that all this while your weight was being balanced by the normal reaction from the surface on which you were standing. Also, as only 2 forces were ...


0

The spring is not external to the system of two gliders. The total momentum is conserved.


1

Model the ground as massless critically-damped vertical spring that the particle contacts at zero height. When the particle reaches zero height, it has some KE which is dissipated by the damping mechanism. When in contact with the spring, there are three forces acting on the particle, gravity downward and the damper and spring force upward. The net force ...


1

When an object falls and hits the ground - which forces are involved to change its momentum? Vectorial sum of all the forces acting on the object will cause the change in momentum of the object. When the object was in free-fall, its momentum was already changing due to gravity(assuming negligible amount of air resistance) and then it hit the ground. ...


1

In Newtonian Mechanics, if a body of mass $\mathtt{m}$ is in free-fall, then gravitational force is responsible for acceleration & hence changing its momentum. Simple, right? The equation of motion is $$\mathtt{m}\cdot a = \mathbf{F_g} = \mathtt{m} \cdot g$$ where $a$ is the net acceleration of the body. Things become intricate when you consider a ...


1

In this cases, momentum is not conserved because of the action of gravity as an external force. When you have a pivoted rod, as in your problem, you can use basically two conservation laws: a) conservation of energy, if the collision is assumed as perfectly elastic; b) conservation of angular momentum about the pivot. As regards b), indeed, if we choose ...


2

In a 1D elastic collision, it is well-known that the relative velocities of the two objects (before and after the collision) are reversed. When you say reversed do you mean that each object keeps their own velocity just with a change of sign? That would not always be the case for a 1D situation. If a resting block $v_{1,before}=0$ is hit by a moving ...


1

The symbol $s_{NN}$ is in OP's context of RHIC the Mandelstam $s$-variable in a Nucleus+Nucleus collision. The $s$-variable is also known as the square of the center-of-mass energy.


-2

I think that the law of conservation will hold good in this situation because there is no external force acting on the system. Because like you said the impulsive force by the hinge is internal and no other force is a acting on the system.


-1

If I've interpreted your question correctly - the ball will collide with the rod at the opposite end to the hinge. This will lose energy via usual mechanisms. The rod will then have an instantaneous velocity, hence momentum, and will swing round the hinge. The ball will career off in whatever direction with the remaining momentum resulting from the ...


1

It basically means that they just need to cover half the distance. So, you have the distance to be covered, initial velocity(40) and final velocity has to be zero. Finding deceleration won't be an issue.


0

You asked: Why do both vehicles experience the same magnitude of force? The larger principle at work is conservation of momentum. (Noether's Theorem, symmetry, and all that jazz.) During the small time frame of the collision we generally assume that there is no transfer of momentum into or out of the system of the car and truck. Changes of momentum ...


0

The angular momentum of the system is the same before and after the collision. Since one object is stationary before the collision, the angular momentum is just the momentum ($mv$ of the moving puck multiplied by the perpendicular distance between them (which is $2r\sin(30)$). The moment of inertia of the two pucks stuck together is a little bit tricky, ...


0

A key idea is that the path of the center of mass is unaffected whether the two pucks collide or not. So take the snapshot of the situation at any moment of time before collision. Find the angular momenta of the two pucks with respect to the com at that instant of time. Since angular momentum is conserved for an isolated system (i.e $\tau_{net}=0$), the ...



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